Fitting a function given a set of data points

Question

I'm trying to fit a fuction given a set of data points. My code is:

f[T_?NumericQ] := NIntegrate[ y^5/((1 - Exp[-y]) (Exp[y] - 1)), {y, 0, (d/T)}]

FindFit[data, (T^5/d^4) f[T], {d}, T]

But it returns me this error:

FindFit::fmgz: Encountered a gradient that is effectively zero. The result returned may not be a minimum; it may be a maximum or a saddle point.

Could anyone please explain me what it means and how to solve it?

Yes, it's a code to find the Debye temperature of a metal after measuring resistivity at different temperatures — Mario Papace, May 14 '20 at 15:03
See, you should have put things like that in your question... — J. M.'s missing motivation, May 14 '20 at 15:05
Have you tried using methods other than the automatic method? Also, the NonlinearModelFit function might be more useful in the long run for analyzing your fits. — BioPhysicist, May 14 '20 at 15:10
Can you share your data? I assume that your data are the ones shown here: https://mathematica.stackexchange.com/questions/221824/fitting-of-an-integral-function-given-a-set-of-data-points. Please try to share them in a "copypastable" form. — demm, May 14 '20 at 15:31
You should also plot (T^5/d^4) f[T] over the range of your data for various values of d. I think you'll find that the proposed function is inadequate to summarize your data. — JimB, May 14 '20 at 16:48
Yes, I also tried to use the NonlinearModelFit function but it returns me the same error. — Mario Papace, May 14 '20 at 18:48
Sorry I don't know how to share my data, in the comment section as well as in he answer section it doesn't allow you to insert text files, moreover I can't just make a copypaste because characters exceed the maximum allowed. — Mario Papace, May 14 '20 at 18:49
The function is a good approximation to my data, I also fitted them using python and the fit is good, d is supposed to be the fit parameter. — Mario Papace, May 14 '20 at 18:52
@MarioPapace I wasn't saying that to suggest it will fix your error. I am saying if you want to be able to do more analysis on your fits then that is a better function to use once you get the fitting to work — BioPhysicist, May 14 '20 at 18:56

score 4 · Answer 1 · answered May 14 '20 at 21:27

I suspect you must have used a different function when you found an acceptable fit with Python.

Here is your data OCR'd from your previous question:

data = {{6.806883, -4.95593692818423 Exp[-41]}, {6.813879, 0.2966197}, {6.81548, 0.2090413}, 
  {6.815993, 0.1047412}, {6.816116, 0.2172217}, {6.816141, 0.0175237}, {6.816558, 0.0401401}, 
  {6.820772, 0.1076685}, {6.831049, 0.273381182476389}, {6.832778, 0.0872403214012297}, 
  {6.835968, 0.213306278687343}, {6.837872, 0.196895932533893}, {6.839626, 0.182350945974245}, 
  {6.842643, 0.236392340715514}, {6.843521, 0.385921906058286}, {6.843593, 0.44360591444177}, 
  {6.845431, 0.395308588265182}, {6.848036, 0.347737788675773}, {6.85891, 1.2787088},
  {6.861215, 1.6855233}, {6.87505, 2.5303902}, {6.888942, 3.59352405445488}, 
  {6.893837, 3.64575756470423}, {6.895161, 3.68408863938315}, {6.89668, 3.66514216133916}, 
  {6.897294, 3.67968886871024}, {6.903002, 3.70207215134901}, {6.908976, 3.67924971763244}, 
  {6.911722, 3.72623544133449}, {6.914223, 3.74474620992355}, {6.917128, 3.75346349648285}, 
  {6.927654, 3.79135369927254}, {6.930277, 3.76666521763759}, {6.93606,  3.8084950461229}, 
  {6.9368, 3.66412103183458}, {6.948534, 3.9201313112154}, {6.949261, 3.81452649019695}, 
  {6.956995, 3.84714584756615}, {6.959176, 3.83848603610872}, {6.968645, 3.85332287228747}, 
  {6.971786, 3.85285653238902}, {6.972535, 3.8534977067291}, {6.981893, 3.86037751083982}, 
  {6.98856, 3.8623598856111}, {6.997314, 3.87119143404958}, {7.013482, 3.88232714899738}, 
  {7.018697, 3.87935255435361}, {7.033174, 3.88806880842605}, {7.058705, 3.89906031937574}, 
  {7.061641, 3.90162364008693}, {7.066072, 3.90278794110276}, {7.080686, 3.9021178571306},
  {7.088082, 3.90925888041691}, {7.089749, 3.90931704384336}, 
  {6.795619, -1.74834441633166 Exp[-41]}};
ListPlot[data]

Running

f[T_?NumericQ] := NIntegrate[y^5/((1 - Exp[-y]) (Exp[y] - 1)), {y, 0, (d/T)}]
FindFit[data, (T^5/d^4) f[T], {{d, 0.01}}, T]

does result in the error you mention (and one other error you didn't mention).

Because there is only a single parameter to fit we can find a value of $d$ that minimizes the root mean square error:

rmse[d_] := Sqrt[(1/(Length[data] - 1)) Total[
    Table[(data[[i, 2]] - (data[[i, 1]]^5/d^4) NIntegrate[y^5/((1 - Exp[-y]) (Exp[y] - 1)), 
    {y, 0, (d/data[[i, 1]])}])^2, {i, Length[data]}]]]

A plot of rmse over values of d:

Plot[rmse[d], {d, 0.01, 20}]

What does the fitted function look like with $d=0.01$ ?

Show[ListPlot[data], 
 Plot[(t^5/0.01^4) NIntegrate[y^5/((1 - Exp[-y]) (Exp[y] - 1)), {y, 0, (0.01/t)}], {t, 6.8, 7.1}]]

BioPhysicist · Answer 2 · 2020-05-14T19:23:42.707

I think the issue is that Mathematica is trying to take derivatives of your fitting form, but your fitting form involves NIntegrate, a numerical process. One thing you should be able to do for now to get around this is specify the fitting to use the "NMinimize" method. This method will not try to find gradients of the form you are trying to fit to.

FindFit[..., Method->NMinimize]

NMinimize also has further methods you can specify, outlined in more detail here. Additionally, this answer describes how to implement the various methods in the context of fitting.

Also as I mentioned in the comments, I would suggest using the NonlinearModelFit function rather than the FindFit function. The latter only returns the parameter values, the former gives you a fitting object that can give you the parameters as well as a bunch of additional analytics of the fits.

Fitting a function given a set of data points

2 Answers2